postgresql - Inner join & outer join; is the order of tables in from important?

Question

Why is the order of tables important when combining an outer & an inner join ? the following fails with postgres:

SELECT grp.number AS number,     
       tags.value AS tag   
FROM groups grp,
     insrel archiverel  
LEFT OUTER JOIN ownrel ownrel ON grp.number = ownrel.dnumber   
LEFT OUTER JOIN tags tags ON tags.number = ownrel.snumber   
WHERE archiverel.snumber = 11128188 AND    
      archiverel.dnumber = grp.number 

with result:

ERROR:  invalid reference to FROM-clause entry for table "grp" LINE 5: LEFT OUTER JOIN ownrel ownrel ON grp.number = ownrel.d... 
^ HINT:  There is an entry for table "grp", but it cannot be referenced from this part of the query.

when the groups are reversed in the FROM it all works:

SELECT  grp.number AS number,     
        tags.value AS tag   
FROM    insrel archiverel,
        groups grp
LEFT OUTER JOIN ownrel ownrel ON grp.number = ownrel.dnumber   
LEFT OUTER JOIN tags tags ON tags.number = ownrel.snumber   
WHERE   archiverel.snumber = 11128188 AND    
        archiverel.dnumber = grp.number 

Answer 1

I believe that you can think of this as an operator precedence issue.

When you write this:

FROM groups grp,
     insrel archiverel  
LEFT OUTER JOIN ownrel ownrel ON grp.number = ownrel.dnumber   
LEFT OUTER JOIN tags tags ON tags.number = ownrel.snumber   

I think it is interpreted by the parser like this:

FROM groups grp,
(
  (
     insrel archiverel  
     LEFT OUTER JOIN ownrel ownrel ON grp.number = ownrel.dnumber   
  )
LEFT OUTER JOIN tags tags ON tags.number = ownrel.snumber
)

If so, then in the innermost join "grp" is unbound.

When you reverse the lines with "groups" and "insrel", the innermost join applies to "groups" and "ownrel", so it works.

Probably this would work as well:

    FROM groups grp
         JOIN insrel archiverel  ON archiverel.dnumber = grp.number
    LEFT OUTER JOIN ownrel ownrel ON grp.number = ownrel.dnumber   
    LEFT OUTER JOIN tags tags ON tags.number = ownrel.snumber 
WHERE archiverel.snumber = 11128188